978-0073398198 Chapter 10 Part 2

subject Type Homework Help
subject Pages 14
subject Words 1590
subject Authors Afshin Ghajar, Yunus Cengel

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page-pf1
10-21
page-pf2
10-22
Ts
[F]
h
[Btu/h.ft2.F]
e
W
[kW]
evap
m
260
262
264
266
268
270
272
274
276
278
280
282
284
286
288
290
292
294
296
298
300
12919
18603
25321
33073
41857
51676
62528
74413
87332
101285
116271
132290
149343
167430
186550
206704
227891
250111
273366
297653
322974
4.956
8.564
13.6
20.3
28.9
39.65
52.77
68.51
87.11
108.8
133.8
162.4
194.8
231.2
272
317.2
367.2
422.2
482.4
548.1
619.5
260 265 270 275 280 285 290 295 300
0
50000
100000
150000
200000
250000
300000
350000
0
100
200
300
400
500
600
700
h [Btu/h-ft2-F]
We [kW]
h
We
260 265 270 275 280 285 290 295 300
0
250
500
750
1000
1250
1500
1750
2000
2250
Ts [F]
mevap [lbm/h]
page-pf3
10-23
10-29 Hot mechanically polished stainless steel ball bearings are cooled by submerging them in water at 1 atm. The rate of
heat removed from a ball bearing at the instant it is submerged in the water is to be determined.
75.1Pr
kg/m 5978.0
kg/m 9.957
3
3
=
=
=
l
v
l
KJ/kg 4217
skg/m 10282.0
J/kg 102257
3
3
=
=
=
pl
l
fg
c
h
26
3
W/m101998.2
75.1)102257(0130.0
05890
=
.
The heat transfer surface area is
222 m 007854.0)m 05.0( ===
DAs
The rate of heat removed from a ball bearing at the instant it is submerged in the water is
page-pf4
10-24
10-30 A long hot mechanically polished stainless steel sheet is being cooled in a water bath. The temperature of the
stainless steel sheet leaving the water bath is to be determined whether or not it has the risk of thermal burn hazard.
Assumptions 1 Steady operating conditions exist. 2 Surface temperature is uniform. 3 The boiling regime is nucleate boiling
75.1Pr
=
l
KJ/kg 4217
=
pl
l
c
Also, Csf = 0.013 and n = 1 for the boiling of water on a
mechanically polished stainless steel surface (Table 10-3).
Analysis The mass of the stainless steelsheet being conveyed
enters and exits the water bath at a rate of
W10422.1
6
=
With ΔTexcess = 25°C, nucleate boiling would occur in the water bath. The heat flux can be determined from Rohsenow
relation to be
26
3
3
1/2
33
3
sat,
2/1
nucleate
W/m101998.2
75.1)102257(0130.0
)100125(4217
05890
)597809957(819
Pr
)(
)(
=
=
.
...
hC
TTc
g
hq n
lfgsf
slp
vl
fgl
The heat transfer surface area of the sheet submerged in the water bath is
2
m 01.1)m 005.0)(m 1(2)m 5.0)(m 1(2 =+=
s
A
The rate of heat that could be removed from the sheet in the water bath is
W10422.1) W/m101998.2)(m 01.1( 6
removed
262
nucleateboiling ==== QqAQ s
W102.222 6
page-pf5
10-25
page-pf6
10-26
page-pf7
10-27
page-pf8
10-28
10-34 Water is boiled at a temperature of Tsat = 160°C by a 3 m × 3 m horizontal flat heater heated by hot gases flowing
through an array of tubes embedded in it. The maximum rate of vaporization is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible.
heater)flat large thusand 27 > * (since 149.0
27 >1309
0466.0
)256.34.907(81.9
)3(
)(
*
2/1
2/1
LC
g
LL
cr
vl
=
=
=
=
Then the maximum heat flux is determined from
26
4/123
4/12
max
W/m105252.2
)]256.34.907()256.3(81.90466.0)[102083(149.0
)]([
=
=
= vlvfgcr ghCq
The heat transfer surface area is
2
m 9m 3 m 3 === LLAs
Then, the rate of heat transfer during nucleate boiling becomes
W102727.2) W/m105252.2)(m 9( 7262
maxboiling === qAQ s
The maximum rate of vaporization of water is determined from
kg/s 10.9=
==
J/kg 102083
J/s 102727.2
3
7
boiling
vapor
fg
h
Q
m
page-pf9
10-29
page-pfa
10-30
page-pfb
10-31
page-pfc
10-32
page-pfd
10-33
10-39 A 10 cm × 10 cm flat heater is used for vaporizing refrigerant-134a at 350 kPa. The surface temperature of the heater is
given as 25°C and the heater is subjected to a heat flux of 0.35 MW/m2. The coefficient Csf is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater are negligible. 3 The boiling regime is
802.3Pr
=
l
N/m 0.01084
KJ/kg 1358
=
=
pl
c
Also, n = 1.7 for the boiling of R134a is given.
AnalysisThe heat flux for nucleate boiling can be expressed using the Rohsenow relation to be
3
sat,
2/1
nucleate Pr
)(
)(
lfgsf
slp
vl
fglhC
TTc
g
Using the Rohsenow relation to solve for Csf yields
0.00772=
=
=
3/1
6
34
6/1
7.13
3/1
nucleate
6/1
sat,
1035.0
)108.194)(10589.2(
01084.0
)12.171278(81.9
)802.3)(108.194(
)525(1358
)(
Pr
)(
q
h
g
h
TTc
Cfgl
vl
n
lfg
slp
sf
For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be
heater)flat large thusand 27 > * (since 149.0
27 >8.106
01084.0
)12.171278(81.9
)1.0(
)(
*
2/1
2/1
LC
g
LL
cr
vl
=
=
=
=
The maximum heat flux in the nucleate boiling regime can be determined from
4/123
4/12
max
)]([
= vlvfgcr ghCq
page-pfe
10-34
page-pff
10-35
10-41 Water is boiled at a temperature of Tsat = 155C by hot gases flowing through a mechanically polished stainless steel
pipe submerged in water whose outer surface temperature is maintained at Ts = 160C. The rate of heat transfer to the water,
the rate of evaporation, the ratio of critical heat flux to current heat flux, and the pipe surface temperature at critical heat flux
page-pf10
10-36
10-42 Steam is generated by a 1 m × 1 m flat heater boiling water at 1 atm with an excess temperature above 300°C. The
range of the steam generation rate that can be achieved by the heater is to be determined.
0.0589 N/m (Tables 10-1) and, from Table A-9, ρl = 957.9 kg/m3, ρv = 0.5978 kg/m3, hfg = 2257 × 103 J/kg.
Analysis For a horizontal flat heating element, the coefficient Ccr is determined from Table 10-4 to be
heater)flat large thusand 27 > * (since 149.0
27 >4.399
0589.0
)5978.09.957(81.9
)1(
)(
*
2/1
2/1
LC
g
LL
cr
vl
=
=
=
=
page-pf11
10-37
10-43 Water is boiled at Tsat = 90C in a brass heating element. The surface temperature of the heater is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the heater and the boiler are negligible.
Properties The properties of water at the saturation temperature of 90C
96.1Pr
CJ/kg 4206cN/m 0608.0
=
==
l
pl
Also,
=
sf
C
0.0060 and n = 1.0 for the boiling of water on a brass heating (Table 10-3).
Heating element
page-pf12
10-38
page-pf13
10-39
10-45 The power dissipation per unit length of a metal rod submerged horizontally in water, when electric current is passed
through it, is to be determined.
Assumptions 1 Steady operating condition exists. 2 Heat losses from the boiler are negligible.
10-6. Therefore, film boiling will occur. The film boiling heat flux in this case can be determined from
25
4/1
5
33
sat
4/1
sat
sat
3
filmfilm
W/m10152.1
)400(
)400)(002.0)(10045.2(
)]400)(1997(4.0102257)[3831.09.957)(3831.0()04345.0(81.9
62.0
)(
)(
)](4.0)[(
=
+
=
+
=
TT
TTD
TTchgk
Cq s
sv
spvfgvlvv
The radiation heat flux is determined from
24444284
4
page-pf14
10-40
10-46E Water is boiled at 1 atm pressure and thus at a saturation (or boiling) temperature of Tsat = 212F by a horizontal
polished copper heating element whose surface temperature is maintained at Ts = 788F. The rate of heat transfer to the water
A-16E)
hlbm/ft 0.04561slbm/ft 10267.1
lbm/ft 02571.0
5
3
==
=
v
v
2
4/1
32
sat
4/1
sat
3
film
ftBtu/h 600,18
)212788(
)212788)(12/5.0)(04561.0(
)]212788(4707.04.0970)[02571.082.59)(02571.0()02267.0()3600(2.32
62.0
)(
)(
)](4.0)[(
62.0
=
+
=
+
=TT
TTD
TTchgk
qs
sv
satspvfgvlvv
The radiation heat flux is determined from
 
2
44428
4
sat
4
rad
ftBtu/h 7.761
R) 460212(R) 460788()RftBtu/h 101714.0)(2.0(
)(
=
++=
=
TTq s

Note that heat transfer by radiation is very small in this case because of the low emissivity of the surface and the relatively
low surface temperature of the heating element. Then the total heat flux becomes
2
radfilmtotal ftBtu/h 171,197.761
4
3
600,18
4
3=+=+= qqq
Finally, the rate of heat transfer from the heating element to the water is determined by multiplying the heat flux by the heat
transfer surface area,
Btu/h 2509=
=
==
)ftBtu/h ft)(19,171 1ft 12/5.0(
)(
2
totaltotaltotal
qDLqAQ s
P = 1 atm
Water, 212F

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